Answer:
The new volume of the balloon when the pressure equalised with the pressure of the atmosphere = 494 L.
The balloon expands by am additional 475 L.
Explanation:
Assuming Helium behaves like an ideal gas and temperature is constant.
According to Boyle's law for ideal gases, at constant temperature,
P₁V₁ = P₂V₂
P₁ = 26 atm
V₁ = 19.0 L
P₂ = 1 atm (the balloon is said to expand till the pressure matches the pressure of the atmpsphere; and the pressure of the atmosphere is 1 atm)
V₂ = ?
P₁V₁ = P₂V₂
(26 × 19) = 1 × V₂
V₂ = 494 L (it is assumed the balloon never bursts)
The new volume of the balloon when the pressure equalised with the pressure of the atmosphere = 494 L.
The balloon expands by am additional 475 L.
Hope this Helps!!!
C is the answer.
The temperature T<span> in degrees Celsius (°C) is equal to the temperature </span>T<span> in Kelvin (K) minus 273</span>°.
So you have evidence that the experiment is true or correct.
From the calculations, the concentration of the acid is 0.24 M.
<h3>What is neutralization?</h3>
The term neutralization has to do with a reaction in which an acid and a base react to form salt and water only.
We have to use the formula;
CAVA/CBVB = NA/NB
CAVANB =CBVBNA
The equation of the reaction is; 2NaOH + H2SO4 ----> Na2SO4 + 2H2O
CA = ?
CB = 1.2 M
VA = 50 mL
VB = 20 mL
NA = 1
NB = 2
CA = CBVBNA/VANB
CA = 1.2 M * 20 mL * 1/ 50 mL * 2
CA = 0.24 M
Learn more about neutralization:brainly.com/question/27891712
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Explanation:
Contributing structures are the resonating structures which are formed due to the delocalization of electrons in a molecule.
The azide ion that is 
, is a symmetrical ion, all of whose contributing structures have formal charges.
Lone pair of central nitrogen atom in azide ion is in conjugation with the neighboring nitrogen atoms.
Contributing structures of azide ion are drawn in the image attached.
